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sattari [20]
3 years ago
12

What is the interquartile range of the data set? 5,6,7,8,9,10,11

Mathematics
1 answer:
sasho [114]3 years ago
3 0
The answer that I got was 4 for the interquartile range of the numbers. hope this helped and let me know if the answer is correct.
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Graph the equation: x = y^2 - 2
Lesechka [4]
Draw a cartesian plane, create a graph with the equation x = y^2 - 2
then substitute numbers into the equation so that it is true, to find points on the graph, e.g. substitute y with 1, you get 
x = 1^2 - 2
x = 1 - 2 = -1, so when y = 1, x = -1, this point is (-1, 1)

for the next substitute y with 2,
x = 2^2 - 2
x = 4 - 2 = 2, the point is (2, 2)
you might want to try negative values of y 
y = -1, x = (-1)^2 - 2
x = -1 the point is (-1,-1)

then plot the points on the graph
5 0
3 years ago
Points F (0,0), G (9,0) and H (9,12) of vertical triangle
ludmilkaskok [199]

Answer:

15 is the distance

Step-by-step explanation:

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3 years ago
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Simora [160]
For the domain
3-x >= 0
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7 0
3 years ago
Babies born after 40 weeks gestation have a mean length of 52.2 centimeters (about 20.6 inches). Babies born one month early hav
Sveta_85 [38]

Answer:

a) Z = -2.88

b) Z = -0.96

c) 40 weeks gestation babies

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

a. Find the standardized score (z-score), relative to all U.S. births, for a baby with a birth length of 45 cm.

Here, we use \mu = 52.2, \sigma = 2.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{45 - 52.2}{2.5}

Z = -2.88

b. Find the standardized score of a birth length of 45 cm. for babies born one month early, using 47.4 as the mean.

Here, we use \mu = 47.4, \sigma = 2.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{45 - 47.4}{2.5}

Z = -0.96

c. For which group is a birth length of 45 cm more common?

For each group, the probability is 1 subtracted by the pvalue of Z.

Z = -2.88 has a lower pvalue than Z = -0.96, so for Z = -2.88 the probability 1 - pvalue of Z will be greater. This means that for 40 weeks gestation babies a birth length of 45 cm is more common.

3 0
3 years ago
A rectanglelar field has an area of 2100 square feet. The length of the field is 50 feet​
Setler [38]

Answer: 42 feet

Step-by-step explanation:

P-area of field

P=2100sqf

a-length of field

a=50 feet

P=a*b

2100sqf=50*b

b=2100/50

b=42 feet

5 0
3 years ago
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